# I Is subtraction an operation on Z?

#### opus

Gold Member
An operation * on A is a rule which assigns to each ordered pair (a,b) of elements of A exactly one element a*b in A.
There are 3 criterias that need to be met:
1) a*b is defined for every ordered pair (a,b) of elements of A.
2) a*b must be uniquely defined (unambiguous).
3) If a and b are in A, a*b must be in A.

Trying to figure out if subtraction meets these criterias on the integers. I know that we define subtraction in terms of addition, but I'm trying to go at this from scratch.

Now I have a couple questions that I'm unsure of.
For criteria 1: a*b needs to be defined for every (a,b) in A. But defined in terms of what? Axioms? If so, what axioms?
For criteria 2: Im not sure how to determine if this is uniquely defined. I know that a-b is unambiguous, in the sense of our daily use of it. But I'm not sure if I should be thinking about it that way.

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#### fresh_42

Mentor
2018 Award
An operation * on A is a rule which assigns to each ordered pair (a,b) of elements of A exactly one element a*b in A.
There are 3 criterias that need to be met:
1) a*b is defined for every ordered pair (a,b) of elements of A.
2) a*b must be uniquely defined (unambiguous).
3) If a and b are in A, a*b must be in A.

Trying to figure out if subtraction meets these criterias on the integers. I know that we define subtraction in terms of addition, but I'm trying to go at this from scratch.
Indeed. Subtraction doesn't exist, because $a-b = a+ (-b)$ is usually only an abbreviation.
Now I have a couple questions that I'm unsure of.
For criteria 1: a*b needs to be defined for every (a,b) in A. But defined in terms of what? Axioms? If so, what axioms?
$\forall\,a\in A \,\,\forall\,b\in A \,\,\exists \,a-b$
For criteria 2: Im not sure how to determine if this is uniquely defined. I know that a-b is unambiguous, in the sense of our daily use of it. But I'm not sure if I should be thinking about it that way.
$c\in A \, : \,c=a-b \,\wedge \,d\in A \, : \,d=a-b \Longrightarrow c=d$

#### opus

Gold Member
∀a∈A∀b∈A∃a−b
Ok I can see how this is true.

c∈A:c=a−b∧d∈A:d=a−b⟹c=d
So here I believe you are saying that if c and d are in the set A, and if c=a-b and d=a-b, then c=d and therefore a-b is unique?
Also, this would clearly be a closed operation.

But then it seems like all 3 criteria have been met and subtraction would be an operation? Of course there's the commutativity issue, but those arent considered in the criteria for an operation.

#### fresh_42

Mentor
2018 Award
But then it seems like all 3 criteria have been met and subtraction would be an operation? Of course there's the commutativity issue, but those arent considered in the criteria for an operation.
Of course it is an operation. Just not what is usually meant by subtraction as the addition of an inverse element.

#### opus

Gold Member
Ahh ok I see. Thank you

#### WWGD

Gold Member
Basically, $\mathbb Z$ is a ring, which means it is a group under addition ( implying that addition of any two terms is defined), meaning that every element r has an additive inverse (-r). So $a+(-b)$ is just addition of a with the additive inverse of b, (-b), so $a+(-b)$ is just addition, which is well-defined. Just to repeat what Fresh said since this helps nit sink in. Subtraction is addition by/with the additive inverse, which is guaranteed to exist in a ring, since the ring is a group under addition; every element in a group has an inverse.

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#### opus

Gold Member
Thank you kindly. Just getting used to the abstractness of the subject so sometimes it helps to have it spelt out for me.

#### fresh_42

Mentor
2018 Award
Thank you kindly. Just getting used to the abstractness of the subject so sometimes it helps to have it spelt out for me.
You defined $-$ as an operation on $\mathbb{Z}$. This can be done, but it is neither commutative nor associative.

The usual subtraction is an abbreviation for adding the inverse, and as such it is commutative and associative:
$a-b= a+(-b)=(-b)+a$ and $(a-b)-c = (a+(-b))-c = (a+(-b))+(-c)=a+((-b)+(-c))=a+(-b-c)$.

In this sense, it makes an important difference, whether we define subtraction as stand alone operation, or as the abbreviation we usually associate it with.

#### opus

Gold Member
Understood. Thank you!

#### pasmith

Homework Helper
An operation * on A is a rule which assigns to each ordered pair (a,b) of elements of A exactly one element a*b in A.
There are 3 criterias that need to be met:
1) a*b is defined for every ordered pair (a,b) of elements of A.
2) a*b must be uniquely defined (unambiguous).
3) If a and b are in A, a*b must be in A.
What you have there is the definition of a function from $A \times A$ to $A$. And that's all an operation on $A$ is: a function from $A \times A$ to $A$ which is written with infix notation ($a \cdot b$) rather than normal function notation ($f(a,b)$).

#### FactChecker

Gold Member
2018 Award
Now I have a couple questions that I'm unsure of.
For criteria 1: a*b needs to be defined for every (a,b) in A. But defined in terms of what? Axioms? If so, what axioms?
Defined any way you want to define it. Suppose you define the operation '-' as the third member of a set of ordered triples, (a, b, a-b). Then that is defined for all integers a and b. That is enough.
For criteria 2: Im not sure how to determine if this is uniquely defined. I know that a-b is unambiguous, in the sense of our daily use of it. But I'm not sure if I should be thinking about it that way.
For any two integers, a and b, the calculation a-b is uniquely defined. That is enough.

You could build up the entire theory of addition and subtraction of integers, but I do not think that would be required here and it would be a distraction. The point is to specify the properties that one would want before '*' is called an operation. In the case of subtraction of integers, the fact that it is an operation on the integers is easy to see.